What is the relationship between redshift and conformal time ?

For example in a paper i found:

taking $z_e = 3234 $ at the time of radiaton-matter equality yields the conformal time $\frac{\eta_e}{\eta_0} = 0.007$ and taking $z_E=0.39$ at matter-$\Lambda$ equality yields $\frac{\eta_E}{\eta_0}=0.894$ and setting redshift at decoupling $z_d=1089$ yields $\frac{\eta_d}{\eta_0} = 0.0195$ where $\eta_0$ is the present decoupling time.

Further some cosmological parameters are given as : $\Omega_r = 8.36 \times 10^{-5}, \Omega_m = \Omega_b + \Omega_{dm} = 0.044 +0.226, \Omega_\Lambda = 0.73, H_0=0.72$

Now how can i calculate all those $\eta_e, \eta_E, \eta_d, \eta_0$ from given redshift values and/or above parameters ? I searched whole of my text books trying to find an explicit relation for conformal time $\eta$ but all i got was $ \mathrm{d}t=a(t)\mathrm{d}\eta$. Any help would be very helpful.

  • $\begingroup$ Have you taken a look at this wikipedia page? $\endgroup$
    – Danu
    Oct 29, 2014 at 8:42
  • $\begingroup$ Well yeah. Including that link and almost everywhere, i find relations are given in terms of time "t" rather than conformal time. Also I can find redshift from given fractional density parameters but from redshift to conformal time ... no idea till now. $\endgroup$
    – cmbfast
    Oct 29, 2014 at 9:12

1 Answer 1


As you state, conformal time is defined as $$ \eta(t) = \int_0^t\frac{\text{d}t'}{a(t')}. $$ Using $$ \dot{a} = \frac{\text{d}a}{\text{d}t}, $$ this can be written in the form $$ \eta(a) = \int_0^a\frac{\text{d}a}{a\dot{a}} = \int_0^a\frac{\text{d}a}{a^2H(a)}, $$ with $$ H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}. $$ The scale factor $a$ is related to the redshift as $$ 1 + z = \frac{1}{a}, $$ so that $$ \eta(z) = \frac{1}{H_0}\int_0^{1/(1+z)}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}. $$ Basically, the conformal time is equal to the distance of the particle horizon, divided by $c$ (see this post for more info). $\eta_0$ refers to the current conformal age of the universe.


I just checked the values that you posted with my own cosmology calculator. I get $$ \eta_0 = 45.93\;\text{Gigayears} $$ for the current conformal age of the universe, and $$ \begin{align} z_e &= 3234,& a_e &= 0.000309,& t_e &= 5.54\times 10^{-5}\;\text{Gy},&\eta_e &= 0.3804\;\text{Gy},\\ z_E &= 0.39,& a_E &= 0.719,& t_E &= 9.359\;\text{Gy},& \eta_E &= 41.08\;\text{Gy},\\ z_d &= 1089,& a_d &= 0000917,& t_d &= 0.00037\;\text{Gy},& \eta_d &= 0.911\;\text{Gy}, \end{align} $$ so that $$ \frac{\eta_e}{\eta_0} = 0.00828,\quad \frac{\eta_E}{\eta_0} = 0.894,\quad \frac{\eta_d}{\eta_0} = 0.0198. $$ So my results are almost the same, but there's a small discrepancy. Apparently there's a small numerical error somewhere.

  • $\begingroup$ Wow thanks indeed for clear answer and the link as well ! $\endgroup$
    – cmbfast
    Oct 29, 2014 at 9:55
  • $\begingroup$ wait !! sorry for being lame here but I put the values of $z$ and all $\Omega$s in the above equation but unable to get values as you obtained. I searched few web based cosmology calculators but they dont have $\eta$. Where did i go wrong ? $\endgroup$
    – cmbfast
    Oct 29, 2014 at 10:54
  • $\begingroup$ @cmbfast Did you convert $H_0^{-1}$ into years? $\endgroup$
    – Pulsar
    Oct 29, 2014 at 11:04
  • $\begingroup$ Well yeah just figured that and corrected it and almost got your answers for 1st and last case. However putting z=0.39 is giving a negative $\eta$ . $\endgroup$
    – cmbfast
    Oct 29, 2014 at 11:17
  • $\begingroup$ @cmbfast Then there must be a bug in your program. The integral should always be positive. $\endgroup$
    – Pulsar
    Oct 29, 2014 at 11:25

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